Catenary rings Anna Bot Bachelor thesis supervised by Prof. Dr. Richard Pink 2.10.2017 Contents 1 Introduction 1 2 Prerequisites 2 3 Regular sequences 3 4 Depth 5 5 Depth in light of height, localisation and the Jacobson radical 9 6 Cohen-Macaulay rings 13 7 (Universally) Catenary rings 19 8 Proofs of the main Theorems 22 9 Geometric interpretation 25 References 29 1 Introduction It is assumed that the reader is at ease with the terminology and concepts from Commuta- tive Algebra, for example prime ideals, noetherian rings, quotient rings, local rings, finitely generated algebras over a ring and localisation of rings, to name a few. For an orientation, one might find the lecture notes of Commutative Algebra [1] useful. For the entire exposition, all rings are commutative and unitary. The aim of this bachelor thesis is to prove the following two theorems: Theorem. Any ring that is finitely generated over a field K, or over Z, or over any Dedekind ring, respectively, is catenary. Theorem. Any integral domain R that is finitely generated over a field K, or over Z, satisfies the following: (i) For all prime ideals p ⊂ R: ht(p) + coht(p) = dim(R). (ii) For all maximal ideals m ⊂ R: ht(m) = dim(R). There are two approaches to proving the first statement; one can work with complexes, namely the Koszul complex, or | as will be pursued here | one uses regular sequences, depth, Cohen Macaulay rings and catenary rings. The path we will take is as follows: After introducing and examining regular sequences, we are able to define the depth of an ideal. Then, we anchor the depth to other con- cepts such as localisation, the height of an ideal, and the Jacobson radical. Subsequently, we define Cohen-Macaulay rings and (universally) catenary rings, and show that Cohen- Macaulay rings are universally catenary. This cumulates in the proof of our first theorem. For the second theorem, one does not necessarily need to understand the previous chapters in detail, and can thus skip to Chapter 8. At the end, we look back on our findings and try to find some meaningful geometric interpretation of them. Nearly all proofs in this thesis imitate the proofs found in the book by I. Kaplansky [4] in chapter 3, pages 84-100. Said book is primarily concerned with modules, to which the notions introduced in this bachelor thesis can be extended. Here, the proofs given by Kaplansky were adapted to rings in order to fit into the scope of this bachelor thesis. The proof about polynomial rings over Cohen-Macaulay rings found in Section 6 is taken from the book by D. Eisenbud [3], as are the proofs of Section 7 and the example of a ring that is not Cohen-Macaulay. They are taken from chapter 18, pages 448, 449, 451-453 and 466. I would like to thank my supervisor Professor Pink for his guidance, the careful cor- rections and his useful comments on structure, content and the mathematical language. I learnt from him that to the reader, legibility and a good order is as important as correct proofs. Furthermore, I am grateful to Sebastian Schlegel Mejia for his thorough reading of one of my drafts and making me aware of some inconsistencies. Also, I am thankful to Oliver Edtmair for the helpful discussion of the last proof. 1 2 Prerequisites We state a few important theorems here that will later be referenced in this thesis. Their proofs can be found in standard textbooks on Commutative Algebra, such as in the book Introduction to Commutative Algebra by M.F. Atiyah and I.G. Macdonald [2], or in the book by Eisenbud [3]. Other uses of smaller, but useful facts will be mentioned as we go along. Definition 2.1. A zero-divisor of a ring R is an element x 2 R such that there exists an element y 2 R r f0g with xy = 0. We shall treat 0 as a zero-divisor. Definition 2.2. The height of a prime ideal p ⊂ R is defined as ht(p) := supf r > 0 j There exist prime ideals p0 $ ::: $ pr = p g: The height of an ideal a ⊂ R is defined as ht(a) := inff ht(p) j a ⊂ p; where p is a prime ideal g: Definition 2.3. The coheight of an ideal a ⊂ R is defined as coht(a) := dim(R=a): Theorem 2.4 (Artin-Tate). Suppose R is noetherian and S a finitely generated R-algebra. If T ⊂ S is an R-algebra such that S is a finitely generated T -module, then T is a finitely generated R-algebra. Theorem 2.5. Let L=K be a field extension which is finitely generated as a ring over K. Then L=K is finite. Theorem 2.6 (Hilbert's Basis Theorem). If R is noetherian, then so is R[X1;:::;Xn] for any n 2 Z>0. Theorem 2.7 (Krull's Principal Ideal Theorem). For any noetherian ring R and any a 2 R which is not a zero-divisor, any minimal prime ideal p above (a) satisfies ht(p) = 1. Theorem 2.8 (Krull's Dimension Theorem). For any noetherian ring R, any ideal a = (a1; : : : ; ar) generated by r > 0 elements and any minimal prime ideal p above a we have ht(p) 6 r. If a 6= (1) then ht(a) 6 r. >0 Theorem 2.9. For any noetherian ring R and any n 2 Z , we have dim(R[X1;:::;Xn]) = dim(R) + n: 2 3 Regular sequences Definition 3.1. A regular sequence is a sequence a1; : : : ; ar 2 R for which the following two conditions hold: (i) For every i = 1; : : : ; r, the image of ai in R=(a1; : : : ; ai−1) is not a zero-divisor. (ii) R=(a1; : : : ; ar) 6= 0. If, in addition, a1; : : : ; ar all lie in an ideal a, we call it a regular sequence in a. We see that R = 0 cannot contain a regular sequence. In fact, the empty sequence is regular if and only if R 6= 0, so we exclude the special case of R = 0 from our discussion. The reader can convince her- or himself that the statements of this chapter still hold, albeit vacuously so. As soon as we prove Lemma 4.3 however, the assumption that R 6= 0 must hold becomes necessary. Remark 3.2. We assume R 6= 0. Let us warm up by proving a few direct observations about regular sequences. Lemma 3.3. For every regular sequence a1; : : : ; ar 2 R, we get the chain of ideals (a1) $ (a1; a2) $ ::: $ (a1; : : : ; ar): Proof. For i = 2; : : : ; r,(a1; : : : ; ai−1) ⊂ (a1; : : : ; ai) is an equality if and only if ai 2 (a1; : : : ; ai−1), or equivalently, if ai is zero in R=(a1; : : : ; ai−1). This is excluded by Definition 3.1 and Remark 2.1. As the conditions for a regular sequence are stated in terms of factor rings, we prove the following claim to be able to better deal with them. Lemma 3.4. Let a; b be ideals of R. Then R=a R ∼= ; b(R=a) a + b P0 where b(R=a) = biri j all bi 2 b; all ri 2 R=a . Proof. Consider the composite ring homomorphism R=a ' : R R=a : b(R=a) ∼ R=a This homomorphism is surjective, hence R= ker(') = b(R=a) by the homomorphism theo- rem. We determine the kernel of '. X r 2 ker(') , r + a 2 b(R=a) , (9 bi 2 b 9 ri 2 R 9 a 2 a : r = a + biri) , r 2 a + b: The conclusion follows. 3 Proposition 3.5. Suppose a1; : : : ; ar 2 R is a sequence and let i 2 f1; : : : ; rg. The sequence a1; : : : ; ar is regular if and only if a1; : : : ai−1 is a regular sequence and ai;:::; ar is a regular sequence in R=(a1; : : : ; ai−1). Proof. If a1; : : : ; ar is a regular sequence, then for any i 2 f1; : : : ; arg, we see that a1; : : : ; ai−1 is a regular sequence, too. Now assume a1; : : : ; ai−1 to be a regular sequence. ∼ Set R := R=(a1; : : : ; ai−1). By Lemma 3.4, we get the isomorphism R=(a1; : : : ; ar) = R=(ai;:::; ar). Hence, for any i 6 j 6 r, the image of aj in R=(a1; : : : ; aj−1) is a non-zero- divisor if and only if the image of aj in R=(ai;:::; aj−1) is a non-zero-divisor. Furthermore, we have R=(a1; : : : ; ar) 6= 0 if and only if R=(ai;:::; ar) 6= 0, which concludes the proof. Example 3.6. Consider the polynomial ring Z[X1;X2;X3] and the regular sequence X2;X1(X2 + 1);X3(X2 + 1): If we change the order to X1(X2 + 1);X3(X2 + 1);X2; we see that X3(X2 + 1) is a zero-divisor of Z[X1;X2;X3]=(X1(X2 + 1)), which implies that the second sequence is not regular. We see from the above example that we are not allowed to arbitrarily permute the elements of a regular sequence. This is the motivation for why we would like to know under what circumstances the sequence that results from a regular sequence when changing the order is again a regular sequence. Lemma 3.7. If a; b 2 R is a regular sequence then the image of a in R=(b) is not a zero-divisor. Proof. Suppose a in R=(b) is a zero-divisor. Then there exists a non-zero z 2 R=(b) such that a z = 0 in R=(b): So, there exists z 2 R r (b) such that az 2 (b), i.e.